new coding scheme to me - is it familiar to you?

Bits

I am trying to figure out this coding scheme. I send the number 1..300 and I get two

Any idea what is going on? Thanks.

ElectricAye

Bits wrote: »

... I send the number 1..300 and I get two bytes....

To what are sending you sending the numbers?

Dave Hein

The number is encoded as two hex digits, where each digit is the hex value + $40. This uses the ASCII characters "@" through "O".

Heater.

Well if you remove the leading "4" digit from each number in square brackets and then remove the spaces within the brackets you get:

[01] = 1
[02] = 2
[03] = 3
[04] = 4
[0A] = 10
[0B] = 11


[7D] = 125
[96] = 150
[AF] = 175
[C8] = 200
[E1] = 225
[FA] = 250
[FF] = 255
[2C] = 300


And here we see the number in brackets is the hexadecimal equivalent of the numbers on the right.

Except for [2C] which is only 44 in decimal.

But I note that 12C hex is 300 decimal. Are you sure that last entry [52 4C]?

mindrobots

"It" is sort of converting them to hex values. I'm not sure why it is putting 0x40 with it. 255 = 0xFF = 0x4F 0x4F
..and the other numbers work out similarly. It's not ASCII, not EBCDIC. It's not carrying or indicating an overflow or a wrap around.

Bits

Heater. wrote: »

Well if you remove the leading "4" digit from each number in square brackets and then remove the spaces within the brackets you get:

[01] = 1
[02] = 2
[03] = 3
[04] = 4
[0A] = 10
[0B] = 11


[7D] = 125
[96] = 150
[AF] = 175
[C8] = 200
[E1] = 225
[FA] = 250
[FF] = 255
[2C] = 300


And here we see the number in brackets is the hexadecimal equivalent of the numbers on the right.

Except for [2C] which is only 44 in decimal.

But I note that 12C hex is 300 decimal. Are you sure that last entry [52 4C]?

Ah I think you nailed it - I bet 255 is max and thus the oddity. Thanks everyone.

Dave Hein

300 is $12C

EDIT: Oh, I see that Heater already mentioned that "12C hex is 300 decimal". So it apears it's only encoding the 8 LSBs of the number.

Circuitsoft

I wonder if it's UTF-7...

Bits

Okay i think this converts it

Pub Numbers(a,b) : out 


  out := (((A - $40) << 4) | (B - $40))

Thanks all for the help

jmg

Dave Hein wrote: »

The number is encoded as two hex digits, where each digit is the hex value + $40. This uses the ASCII characters "@" through "O".

Yes, a Quasi-Hex encoding, but one that overlaps std hex Chars.

The sort of thing you might use if you were desperately short of space to unpack Std Hex , or if you wanted a (little) obfuscation, or maybe if you wanted to know it was your unit at the far end ?

I think the ancient UU64 was quite similar, only it used a 32d base adder, and allowed 6 bit binary fields.

{ We have looked at using an alias-hex scheme here in designs, where we 'duplicate' Hex, but safely away from std hex, to allow control messages to be added to a standard Hex stream. }

jmg

Bits wrote: »

Okay i think this converts it

Does it never send anything apart from 40H..4FH ?

Intel hex uses the "':" as a block start indicator, and can skip CrLf, so I would check if this has other characters for block/control indicators.

Bits

No it just sends what I listed above except the space. Kind of strange but hacking this device I seen worse.

Bits

Now it appears there is a 3rd numbering scheme as well.

Pub Numbers(a,b,c) : out

      out := (A - $40) << 4 | (B - $40)
      out := OUT<<4 |  (C - $40)

SOLVED***